Introduction to signals - Continuous and Discrete Time Signals and Systems - page 18

x ( t ) z ( t )

5 5 5 5

t t t t

−8 −6 −4 0 246 −8 −6 −4 0 246

−2 −2

(a) (b)

Problem 1.6). Similarly, the average power of the complex exponential signal
A exp( j ω 0 t )isgivenby A 2

Fig. 1.9. CT signals for Example
1.6.

(see Problem 1.8).

Example 1.6
Consider the CT signals shown in Figs. 1.9(a) and (b). Calculate the instanta-
neous power, average power, and energy present in the two signals. Classify
these signals as power or energy signals.

Solution
(a) The signal x ( t ) can be expressed as follows:

5 − 2 ≤ t ≤ 2

x ( t ) =

0 otherwise .

The instantaneous power, average power, and energy of the signal are calculated
as follows:

− 2 ≤ t ≤ 2

25

instantaneous power P x ( t ) =

0 otherwise;

∞ 2

x ( t ) 2 d t energy E x = = 25 d t = 100;

−∞ − 2

1
T

average power P x = lim

T →∞

E x = 0 .

Because x ( t ) has finite energy (0 < E x = 100 < ∞ ) it is an energy signal.
(b) The signal z ( t ) is a periodic signal with fundamental period 8 and over
one period is expressed as follows:

5 − 2 ≤ t ≤ 2
02 < t ≤ 4 ,

z ( t ) =

with z ( t + 8) = z ( t ). The instantaneous power, average power, and energy of
the signal are calculated as follows:

25 − 2 ≤ t ≤ 2

instantaneous power P z ( t ) =

< t ≤ 4